The document discusses solving quadratic equations by factoring. It provides examples of factoring quadratic expressions to find the solutions to the equations. These include using the zero product rule, factoring a common factor, and factoring a perfect square. It also provides two word problems involving consecutive integers and the Pythagorean theorem and shows how to set up and solve the quadratic equations derived from the word problems.

1. Solving Quadratic Equations by Factoring Quadratic Equations are also known as Second Degree Equations because the highest power of the variable is 2. They may have zero, one or two solutions. There are several methods for solving them. This lesson involves those that can be solved by factoring.

3. Then determine whether the expression on the left side of the equation can be factored. If so, then the product of those factors is 0, and since(factor1)• (factor2) = 0, then either factor1 = 0, or factor2 = 0.

4. Solve by Factoring Here is an example of an equation in Standard Form: y2 – 6y + 5 = 0 Can we factor the expression on the left side of the equation? Yes, we can express it as the product of two polynomial factors. Since the product of those two factors is zero, then according to the Zero Product Rule, one of the factors must be equal to zero. So we set each of the factors to zero and solve to determine the two possible solutions. y2 – 6y + 5 = 0 (y – 5) (y – 1) = 0 y – 5 = 0 OR y – 1 = 0 y = 5 y = 1

5. Check Substitute each of the solutions into the original equation for y = 5 y2 – 6y + 5 = 0 (5)2 – 6(5) + 5 = 0 25 – 30 + 5 = 0 -5 + 5 = 0 0 = 0 for y = 1 y2 – 6y + 5 = 0 (1)2– 6(1) + 5 = 0 1 – 6 + 5 = 0 -5 + 5 = 0 0 = 0 Both solutions are valid.

6. Solve by Factoring – common factor 9 x2 – 5 = 12x – 5 9x2 = 12x 9x2– 12x = 0 9x2– 12x = 0 3x(3x – 4) = 0 3x = 0 OR 3x - 4 = 0 x = 0 3x = 4 x = In this example, the equation is not in standard form, so the first step is to express the equation in standard form. To do so, we move the terms from the right to the left side of the equation. The constant term cancels out in our resulting equation. Next, we factor the expression on the right side of the equation. We find that we have a common factor of 3x in each of the terms of the expression. Using the zero product rule, we set each of the resulting factors equal to zero and solve to find the two solutions.

7. Check Substitute each of the solutions into the original equation for x = 0 9 x2 – 5 = 12x – 5 9(0)2 – 5 = 12(0) – 5 9(0) – 5 = 12(0) – 5 0 – 5 = 0 – 5 -5 = -5 for x = 9 x2 – 5 = 12x – 5 9( )2 – 5 = 12( ) – 5 9( ) – 5 = 12( ) – 5 16 – 5 = 16 - 5 11 = 11 Both solutions are valid.

8. Solve by Factoring – perfect square This equation is in standard form, so we need to determine whether the expression on the right side of the equation can be factored. First we find that there is a common factor of 3, so we factor it out. The resulting expression can be factored further. We recognize it as a perfect square. We can divide both sides of the equation by 3 and are left with two identical factors of (x + 2). According to the zero product rule, we know that x + 2 must be equal to 0, so we set it to zero and solve. 3x2 + 12x + 12 = 0 3(x2+ 4x + 4) = 0 3(x + 2)2 = 0 (x + 2)2 = 0 x + 2 = 0 x = -2

9. Check Substitute the solution into the original equation for x = -2 3x2 + 12x + 12 = 0 3(-2)2 + 12(-2) + 12 = 0 3(4) + 12 (-2) + 12 = 0 12 – 24 + 12 = 0 0 = 0 The solution is valid

10. Applications with Quadratic equations Consecutive Integer Problem We have three consecutive even integers. The sum of the first two integers is equal to one-fourth the product of the second and third. Find all possible solutions for the three integers. Using FFFSA method: FIND: We need to find three consecutive even integers. We can represent them as x, x +2 and x + 4 FACTS: The sum of the first two is represented by: x + (x + 2) The product of the second and the third is: (x + 2)(x + 4) One fourth of that value is: FORMULA: From the facts we can create an equation: x + x + 2 =

11. Consecutive Integer Problem SOLVE: x + x + 2 = 2x + 2 = combine like terms 4(2x + 2) = (x + 2)(x + 4) multiply both sides by 4 8x + 8 = x2 + 6x + 8 left: distributive property / right: FOIL 8x + 8 – 8 = x2 + 6x + 8 – 8 subtract 8 from both sides 8x = x2 + 6x simplify 0 = x2 + 6x – 8x subtract 8x from both sides 0 = x2 – 2x combine like terms x2 – 2x = 0 switch sides (symmetry principle) x(x - 2)= 0 factor out common factor of x x = 0 OR x – 2 = 0 Set factors equal to zero (zero x = 2 product rule) , and solve ANSWER: If x is 0, the second integer is x+2 = 0+2 = 2, the third integer is x+4 = 0+4 = 4 If x is 2, the second integer is x+2 = 2+2 = 4, the third integer is x+4 = 2+4 = 6 So we express the solutions as 0, 2, 4 OR 2, 4, 6

12. Check the SolutionConsecutive Integer Problem Let’s see if our answers fit into the original problem: We have three consecutive even integers. The sum of the first two integers is equal to one-fourth the product of the second and third. Find all possible solutions for the three integers. The solution 0, 2, 4: 0 + 2 = (2)(4) ÷ 4 2 = 8 ÷ 4 2 = 2 True The solution 2, 4, 6: 2 + 4 = (4)(6) ÷ 4 6 = 24 ÷ 4 6 = 6 True Both solutions are possible.

13. Applications with Quadratic equationsRight Triangle Problem The Pythagorean Theorem states that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. It is represented by the formula: a2 + b2 = c2 Here is an example problem: The hypotenuse of a right triangle is 15 cm. and the larger leg is 12 cm. What is the length of the shorter leg? FIND: the length of the shorter leg. We will represent it as: a FACTS: b = 12, c = 15 FORMULA : a2 + b2 = c2 SUBSTITUTE: a2 + (12)2 = (15)2

14. Right Triangle Problem SOLVE: a2 + (12)2 = (15)2 a2 + 144 = 225 simplify a2 – 81 = 0 subtract 225 from both sides of the equation (a + 9) (a – 9) = 0 Since a2 and 81 are perfect squares, we can factor this as the difference of two squares a + 9 = 0 OR a – 9 = 0 Set each factor equal to 0 (Zero product rule) a = -9 a = 9 Solve. ANSWER: The two solutions are 9 and -9. If you check the solutions in the orignal equation, both with result in a true statement, but do they both make sense in the original problem? Remember, we are looking for the length of a side of a triangle. That length cannot be a negative number, so we can throw out the solution -9. So our final answer is that side a is equal to 9 cm.